abb semiconductors ag key parameters v rrm = 2300 v i favm = 1140 a i fsm = 13.5 ka v f0 = 0.83 v r f = 0.30 m w w doc. no. 5sya 1120 - 01 apr-98 avalanche rectifier diode 5sda 10d2303 features optimized for line frequency rectifiers low on-state voltage, narrow v f -bands for parallel operation self protected against transient overvoltages guaranteed maximum avalanche power dissipation industry standard housing blocking part number 5sda 10d2303 5sda 10d2003 5sda 10d1703 condition v rrm 2300 2000 1700 f = 50 hz t p = 10 ms v rsm 2530 2200 1870 t p = 10 ms t j = 160c i rrm 50 ma v rrm t j = 160c p rsm 70 kw t p = 20 s t j = 45c 50 kw t p = 20 s t j = 160c mechanical data f m mounting force min. 10 kn max. 12 kn a acceleration device unclamped device clamped 50 200 m/s 2 m/s 2 m weight 0.25 kg d s surface creepage distance 30 mm d a air strike distance 20.5 mm
abb semiconductors ag 5sda 10d2303 doc. no. 5sya 1120 - 01 apr-98 abb semiconductors ag fabrikstrasse 3 ch-5600 lenzburg, switzerland telephone +41 (0)62 888 6419 fax +41 (0)62 888 6306 on-state i favm max. average on-state current 1140 a half sine wave, t c = 85c i frms max. rms on-state current 1790 a i fsm max. peak non-repetitive 13.5 ka tp = 10 ms t j = 160c surge current 14.5 ka tp = 8.3 ms after surge: i 2 t limiting load integral 910 10 3 a 2 s tp = 10 ms v r ? 0v 875 10 3 a 2 s tp = 8.3 ms v f0 threshold voltage 0.83 v i f = 1000 - 3000 a t j = 160c r f slope resistance 0.30 m w v f min on-state voltage 1.20 v i f = 1800 a t j = 25c v f max on-state voltage 1.35 v thermal t j storage and operating -40...160c junction temperature range r thjc thermal resistance 80 k/kw anode side cooled junction to case 80 k/kw cathode side cooled 40 k/kw double side cooled r thch thermal resistance case to 16 k/kw single side cooled heat sink 8 k/kw double side cooled for a given case temperature t c at ambient temperature t a the maximum on-state current can be calculated as follows: i = -v + (v ) 4 * f * r * p 2 * f r favm f0 f0 2 f 2 f 2 + * where p = t - t r j max c thjc or p = t - t r j max a thja analytical function for transient thermal impedance: z (t) = r (1 - e ) thjc i -t / i 1 4 i t = ? i 1 2 3 4 r (k/kw) 20.95 10.57 7.15 1.33 t i (s) 0.396 0.072 0.009 0.0044 t [s] 10 -3 10 -2 10 -1 10 0 10 1 2 3 4 5 6 7 2 3 4 5 6 7 2 3 4 5 5 6 2 3 4 5 6 7 0 5 10 15 20 25 30 35 40 45 f m =10...12 kn double side cooling z th [k/kw] i favm (a) p (w) v f0 (v) r f ( w ) t max (c) t c (c) t a (c) r thja (k/kw) r thjc (k/kw) f 2 = 1 for dc current 2.5 for half-sine wave 3.1 for 120el., sine 6 for 60 el., sine
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